iOS Swift: Cannot Manipulate String: Value of type 'String' has no member 'firstIndex' [duplicate]

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP



iOS Swift: Cannot Manipulate String: Value of type 'String' has no member 'firstIndex' [duplicate]



This question already has an answer here:



Hi I am trying to manipulate a String, though the code from Apples developer Site does not work:


let name = "Marie Curie"
let firstSpace = name.firstIndex(of: " ") ?? name.endIndex
let firstName = name[..<firstSpace]
// firstName == "Marie"



You can find the example here: https://developer.apple.com/documentation/swift/string



Seems like firstIndex(of: " ") does not exist any longer


firstIndex(of: " ")



I don't understand why this is not working is there a new func with a different name or what could be the problem? I am confused did Apple missed this one and nobody needed it so far???



Thanks a lot



There is a question that is similar though its answered with index(of: Character) though I am looking for something like index(of: String)



This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.





developer.apple.com/documentation/swift/string/… Check your Xcode version
– rv7284
Aug 6 at 13:53






got version 9.4.1
– SirJoe
Aug 6 at 14:01





See Carlos' answer for a fix along your original intention; if you only want to use this to get first/last name from a string you might also be interested in name.split(separator: " ") which returns ["Marie", "Curie"].
– dr_barto
Aug 6 at 14:16


name.split(separator: " ")


["Marie", "Curie"]





no actually its a little more complex, I could not do it that way, I just took this as an easy example
– SirJoe
Aug 6 at 14:28




2 Answers
2



it all depends on what version of swift you are using, the continuous improvement of the swift language , includes syntax changes



replace firstIndex(of: to .index(of:


firstIndex(of:


.index(of:


let name = "Marie Curie"
let firstSpace = name.index(of: " ") ?? name.endIndex
let firstName = name[..<firstSpace]





Yeah I tried that too but I am getting an error there "Cannot convert value of type 'String' to expected argument type 'Character" so seems like this is only available for Character and not for Strings
– SirJoe
Aug 6 at 14:27






what version of xcode' and swift do you have in your project
– Carlos Andres Chaguendo
Aug 6 at 15:36





Xcode: 9.4.1 and Swift: 4.1
– SirJoe
Aug 6 at 16:32





Well there's your answer @SirJoe. You are looking at the docs for Xcode 10 but you are using Xcode 9.
– matt
Aug 6 at 17:26





ah okay, then what was the way you did it before that (I don't wanna download Xcode 10 if not necessary since its still in beta)
– SirJoe
Aug 6 at 17:48


let string = "Marie Curie"
let firstCharIndex = string.index(string.startIndex, offsetBy: 1)
let firstChar = string.substring(to: firstCharIndex)
print(firstChar)





the request is to search for a string this would not solve the problem in my case since the string I have to manipulate is unknown and could be anything I just know for sure that it contains some string sequence
– SirJoe
Aug 6 at 14:02





While this code may answer the question, providing additional context regarding how and/or why it solves the problem would improve the answer's long-term value.
– Nic3500
Aug 7 at 0:26

-

Popular posts from this blog

Firebase Auth - with Email and Password - Check user already registered

Image
Clash Royale CLAN TAG #URR8PPP Firebase Auth - with Email and Password - Check user already registered I want to check when a user attempts to signup with createUserWithEmailAndPassword() in Firebase user Authentication method, this user is already registered with my app. createUserWithEmailAndPassword() 8 Answers 8 To detect whether a user with that email address already exists, you can detect when the call to createUserWithEmailAndPassword () fails with auth/email-already-in-use . I see that @Srinivasan just posted an answer for this. createUserWithEmailAndPassword () auth/email-already-in-use Alternatively, you can detect that an email address is already used by calling fetchSignInMethodsForEmail() (previous called fetchProvidersForEmail() ). fetchSignInMethodsForEmail() fetchProvidersForEmail() When the user trying to create an user with same email address, the task response will be "Response: The email address is already in use by another account." mFirebas
Read more

Dynamically update html content plain JS

Image
Clash Royale CLAN TAG #URR8PPP Dynamically update html content plain JS So I have a script bound in my index.html, which literally has to fetch an array of jsons from an api, then dynamically update my page when get response. But as a result I'm getting an as the only update of my page. here is my script fetch(url) .then((resp) => resp.json()) .then((resp) => resp.results.forEach(item => fetch(item.url) .then((response)=> response.json()) .then((response) => pokeArr.push(response)) .catch(console.log) ) ).catch(console.log) const filler = () => if(!pokeArr) return 0 pokeArr.map((i,j)=> return `<tr>$i.name</tr>` ) const pokeindex = () => document.getElementById('a').appendChild(document.createElement(filler())) pokeindex() When I'm consoling it, I can see in the console all the responses I get, so I'm at least doing the fetching part right. What's parberakan ? What's pokeArr ? – T.J. Crowder Aug 8 at 12
Read more

How to determine optimal route across keyboard

Image
Clash Royale CLAN TAG #URR8PPP How to determine optimal route across keyboard I'm trying to automate the typing process in a video game. Players are given an on-screen QWERTY keyboard layout that they can navigate with the control stick (up, down, left, right) in order to select a letter to type. I was wondering if there was a module for Python (or some other resource I could explore) that would help my program find a path across the keyboard as it types each letter in a given message. In other words, if the first letter typed was "A" and the next letter was "B", the module would know that the program should move 1 space down on the keyboard and 4 spaces to the right. Currently, I have the program resetting to the "G" key after each letter is typed (allowing me to use fixed-directions as the program parses each character in the desired message). As you can imagine, this is quite inefficient and defeats the purpose of automating the typing in the fir
Read more